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andrezito [222]

3 years ago

6

Lacey measured a neighborhood park and made a scale drawing. The scale she used was 3 millimeters = 1 meter. What is the drawing

's scale factor?

1 answer:

Vikki [24]3 years ago

8 0

Answer:

0.003.

Step-by-step explanation:

We have been given that Lacey measured a neighborhood park and made a scale drawing. The scale she used was 3 millimeters = 1 meter.

To find the scale factor, we will convert 1 meter into millimeters as:

We know that 1 meter equals 1000 millimeters.

$\text{Scale factor}=\frac{\text{Drawing side}}{\text{Original side}}$

$\text{Scale factor}=\frac{3}{1000}$

$\text{Scale factor}=0.003$

Therefore, the scale factor of drawing is 0.003.

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Your firm wants to investigate how much money the typical tourist will spend on their next visit to New York. How many tourists

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Answer: 46

Step-by-step explanation:

Given : Margin of error : E= 17

Standard deviation : $\sigma=70$

Critical value for 90% confidence level : $z_{\alpha/2}=1.645$

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The method of moment (MOM) estimator as: $\mathbf{\hat {\theta} =(\dfrac{\overline X}{1-\overline X})^2}$

$\overline X = \dfrac{4}{9}$

$\mathbf{\hat {\theta} =\dfrac{16}{25} }$

Step-by-step explanation:

From the question, the correct format for the probability density function is:

$fx(x ; \theta) = \left \{ {{\sqrt{\theta x}^{\sqrt{\theta}-1}}\ \ 0 \leq x \leq 1 \atop {0} \ \ \ \ \ \ \ otherwise } \right.$

where θ > 0 is an unknown parameter.

(a) The MOM estimator can be calculated as follows:

$E(X) = \int ^1_0x. \sqrt{\theta} \ x^{\sqrt{\theta}-1} \ dx$

$E(X) = \int ^1_0 \sqrt{\theta} \ x^{\sqrt{\theta}} \ dx$

$E(X) = \dfrac{\sqrt{\theta} }{\sqrt{\theta} +1 } ( x ^{\sqrt{\theta}+1})^1_0$

$E(X) = \dfrac{\sqrt{\theta} }{\sqrt{\theta} +1 }$

suppose E(X) = $\overline X$

Then;

$\overline X = \dfrac{\sqrt{\theta} }{\sqrt{\theta} +1 }$

$\dfrac{1}{\overline X} = \dfrac{\sqrt{\theta} +1 }{\sqrt{\theta}}$

$\dfrac{1}{\overline X} =1 + \dfrac{1}{\sqrt{\theta}}$

making $\dfrac{1}{\sqrt{\theta}}$ the subject of the formula, we have:

$\dfrac{1}{\sqrt{\theta}} =\dfrac{1}{\overline X} - 1$

$\dfrac{1}{\sqrt{\theta}} =\dfrac{1-\overline X}{\overline X}$

$\sqrt{\theta} =\dfrac{\overline X}{1-\overline X}$

squaring both sides, we have:

The method of moment (MOM) estimator as: $\mathbf{\hat {\theta} =(\dfrac{\overline X}{1-\overline X})^2}$

b) If the observations are $\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{2}$

Then,

$\overline X = \dfrac{\dfrac{1}{2}+ \dfrac{1}{3}+\dfrac{1}{2}}{3}$

$\overline X = \dfrac{\dfrac{3+2+3}{6}}{3}$

$\overline X = \dfrac{\dfrac{8}{6}}{3}$

$\overline X = \dfrac{8}{6} \times \dfrac{1}{3}$

$\overline X = \dfrac{8}{18}$

$\overline X = \dfrac{4}{9}$

Finally, the point estimate of the estimator $\theta$ is

$\mathbf{\hat {\theta} =\begin {pmatrix} \dfrac{\dfrac{4}{9}}{1-\dfrac{4}{9}} \end {pmatrix}^2}$

$\mathbf{\hat {\theta} =\begin {pmatrix} \dfrac{\dfrac{4}{9}}{\dfrac{5}{9}} \end {pmatrix}^2}$

$\mathbf{\hat {\theta} =\begin {pmatrix} \dfrac{4}{5} \end {pmatrix}^2}$

$\mathbf{\hat {\theta} =\dfrac{16}{25} }$

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Step-by-step explanation:

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